陶哲轩实分析9.7及习题-Analysis I 9.7

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介值定理。很短但很有用。

Exercise 9.7.1. Prove Corollary 9.7.4.
Corollary 9.7.4.(Images of continuous functions). Let a<b, and let f:[a,b]\to \mathbf R be a continuous functions on [a,b]. Let M:=\sup_{x\in [a,b]}f(x) be the maximum value of f, and let m:=\inf_{x\in [a,b]}f(x) be the minimum value.Let y be a real number between m and M (i.e. m\leq y\leq M). Then there exists a c\in [a,b] such that f(c)=y. Furthermore, we have f([a,b])=[m,M].

Solution: Since f is a continuous function on [a,b], by the maximum principle, there is x_1,x_2\in [a,b] such that

M=f(x_1 ),\quad m=f(x_2 )

If x_1=x_2, then let c=x_1 and the proof is over, assume x_1<x_2, then by Exercise 9.4.6, we have f_{[x_1,x_2]} a continuous function on [x_1,x_2 ], by Theorem 9.7.1, \exists c\in [x_1,x_2 ]\subseteq [a,b], s.t. f(c)=y.

\blacksquare

Exercise 9.7.2. Let f:[0,1]\to [0,1] be a continuous function. Show that there exists a real number x in [0,1] such that f(x)=x. This point x is known as a fixed point of f, and this result is a basic example of a fixed point theorem, which play an important role in certain types of analysis.

Solution: We let F(x)=f(x)-x, by Proposition 9.4.9 and Exercise 9.4.6, F(x) is continuous on [0,1], since f has range [0,1], we know that f(0)\geq 0 and f(1)\leq 1, thus

F(0)=f(0)\geq 0,\quad F(1)=f(1)-1\leq 0

Since F(0)\geq 0\geq F(1), by the Intermediate value theorem, there exists c\in [0,1] such that F(c)=0, or f(c)=c, this is the fixed point we search for.

\blacksquare

陶哲轩实分析9.6及习题-Analysis I 9.6

christangdt留下评论

这一节是最值定理，即闭区间上连续函数一定可取到最大值和最小值。

Exercise 9.6.1. Give examples of

( a ) a function f:(1,2)\to \mathbf R which is continuous and bounded, attains its minimum somewhere, but does not attain its maximum anywhere;
( b ) a function f:[0,\infty)\to \mathbf R which is continuous and bounded, attains its maximum somewhere, but does not attain its minimum anywhere;
( c ) a function f:[-1,1]\to \mathbf R which is bounded but does not attain its minimum anywhere or its maximum anywhere;
( d ) a function f:[-1,1]\to \mathbf R which has no upper bound and no lower bound.

Explain why none of the examples you construct violate the maximum principle.

Solution:
( a ) f(x)=|x-3/2| attains minimum 0 at 3/2, but can’t attain its maximum.
( b ) f(x)=a^x,0<a<1 attains it’s maximum 1 at 0, but can’t attain its minimum.
( c ) f(x)=\begin{cases}0, & x=-1\\x, & x\in (-1,1)\\0, & x=1\end{cases}
( d ) f(x)=\begin{cases}0, & x=0\\ \dfrac{1}{x}, & x\in [-1,0)\cup (0,1]\end{cases}

For the examples in ( a ) and ( b ), the intervals are not closed.
For the examples in ( c ) and ( d ), the functions are not continuous.

\blacksquare

陶哲轩实分析9.5及习题-Analysis I 9.5

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左极限和右极限，相对简单的一节。

Exercise 9.5.1. Let E be a susbet of \mathbf R, let f:E\to \mathbf R be a function, and let x_0 be an adherent point of E. Write down a definition of what it would mean for the limit \lim_{x\to x_0;x\in E}f(x) to exist and equal +\infty or -\infty. If f:\mathbf R\backslash {0}\to \mathbf R is the function f(x):=1/x, use your definition to conclude f(0+)=+\infty and f(0-)=-\infty. Also, state and prove some analogue of Proposition 9.3.9 when L=+\infty or L=-\infty.

Solution: Define \lim_{x\to x_0;x\in E}f(x)=+{\infty} iff \forall M>0,\exists \delta >0, s.t. x\in E,|x-x_0 |<\delta would lead to f(x)>M. \lim_{x\to x_0;x\in E}f(x)=-{\infty} iff \forall M>0,\exists \delta >0, s.t. x\in E,|x-x_0 |<\delta would lead to f(x)<-M.
For f(x)=1/x, we consider f(0+) and f(0-):
For \forall M>0,\exists \delta =1/(M+1)>0, s.t. x\in E\cap (0,+{\infty}),|x-0|<\delta, we have

f(x)=1/x>1/\delta =M+1>M

Thus f(0+)=+{\infty} .
For f(0-), let \forall M>0,\exists \delta =-1/(M+1)<0, s.t. x\in E\cap (-{\infty},0),|x-0|<\delta , we have

f(x)=1/x<1/\delta =-M-1<-M

Thus f(0-)=-{\infty}.

An analogue of Proposition 9.3.9: the following two are equivalent:
(a) \lim_{x\to x_0;x\in E)}f(x)=+{\infty} (or -{\infty} )
(b) \forall (a_n)_{n=0}^{\infty} \in E, \lim_{n\to {\infty} }a_n=x_0, we have \lim_{n\to {\infty} }f(a_n)=+{\infty} (or -{\infty})

Proof :
(a) implies (b):
\forall M>0, since (a) is true, \exists \delta >0, s.t. x\in E,|x-x_0|<\delta would lead to f(x)>M. We have \lim_{n\to {\infty} }a_n=x_0, so \exists N\in \mathbf N, s.t.

|a_n-x_0 |<\delta ,\quad\forall n>N

Then we have f(a_n )>M,\forall n>N, this means \lim_{n\to {\infty} } f(a_n )=+{\infty} .
(b) implies (a):
Let (b) be true, assume \lim_{x\to x_0;x\in E)}f(x)\neq +{\infty} , then we can find a M_0, s.t. for any \delta >0, the set {x\in (x_0-\delta ,x_0+\delta ):f(x)\leq M_0} is non-empty. We use axiom of choice to select

a_n\in {x\in (x_0-1/n,x_0+1/n):f(x)\leq M_0}

Then \lim_{n\to {\infty} }a_n=x_0, but \lim_{n\to {\infty} } f(a_n)\leq M_0<+{\infty}, this contradicts (b).
The case of -{\infty} can be similarly proved.

\blacksquare